[tex]\tan a=-\dfrac{12}5[/tex]
Recall the following identities:
[tex]1+\tan^2a=\sec^2a=\dfrac1{\cos^2a}[/tex]
[tex]\cos^2a=\dfrac{1+\cos2a}a[/tex]
from which we get
[tex]1+\left(-\dfrac{12}5\right)^2=\dfrac1{\cos^2a}[/tex]
[tex]\implies\cos^2a=\dfrac{25}{169}[/tex]
Since [tex]a[/tex] is in the fourth quadrant [tex]\left(\dfrac{3\pi}2<a<2\pi\right)[/tex] we know that [tex]\cos a[/tex] should be positive, so when we take the square root here, we should take the positive root.
[tex]\implies\cos a=\sqrt{\dfrac{25}{169}}=\dfrac5{13}[/tex]
Now recall that
[tex]\cos^2a+\sin^2a=1[/tex]
and since [tex]a[/tex] is in the fourth quadrant, we expect [tex]\sin a[/tex] to be negative. So,
[tex]\sin^2a=1-\cos^2a=\dfrac{144}{169}\implies\sin a=-\sqrt{\dfrac{144}{169}}=-\dfrac{12}{13}[/tex]
One final identity:
[tex]\sin2a=2\sin a\cos a[/tex]
from which we get
[tex]\sin2a=2\cdot\left(-\dfrac{12}{13}\right)\cdot\dfrac5{13}=-\dfrac{120}{169}[/tex]
